Symmetry Point Groups and Topological Entropies of Polyatomic Convex Clusters

Abstract

The H_S topological entropy of all 2907 convex 4- to 9-atomic polyhedral clusters has been calculated from the point of different symmetrical positions of the atoms. It shows a general trend to drop with growing symmetry of clusters with many local exceptions. The H_V topological entropy of the same clusters has been calculated from the point of different valences (chemical bonds) of the atoms. It classifies the variety of clusters in more details. The relationships between the two entropies are discussed.

Appendices

Appendix 1

Consider the sequences of numbers v_i of different valences for convex 5- to 9-acra (related H_V are in parentheses) ordered by the algorithm to follow: … p … q … (H₁) → … p − 1 … q + 1 … (H₂), where 1 ≤ p ≤ q. 5-acra. 23 (0,292) → 14 (0,217). 6-acra. (Hereinafter 0’s and permutations of v_i are omitted in the sequences as they do not affect H_V.) 222 (0,477) → 123 (0,439) → 24 (0,276) → 15 (0,196) → 6 (0). 7-acra. The main trend: 1222 (0,587) → 1123 (0,555) → 223 (0,469) → 133 (0,436) → 124 (0,415) → 34 (0,297) → 25 (0,260) → 16 (0,178); offshoot: 124 (0,415) → 115 (0,346). 8-acra. The main trend: 11123 (0,649) → 1223 (0,574) → 1133 (0,545) → 233 (0,470) → 224 (0,452) → 134 (0,423) → 44 (0,301) → 35 (0,287) → 26 (0,244) → 17 (0,164) → 8 (0); offshoots: 2222 (0,602) → 1223 (0,574); 1133 (0,545) → 1124 (0,527) → 1115 (0,466); and 134 (0,423) → 125 (0,391) → 116 (0,319). 9-acra. The main trend: 11223 (0,661) → 2223 (0,595) → 1233 (0,569) → 333 (0,477) → 234 (0,461) → 144 (0,419) → 135 (0,407) → 45 (0,298) → 36 (0,276) → 27 (0,230) → 18 (0,152) → 9 (0); offshoots: 11223 (0,661) → 11133 (0,636) → 11124 (0,620); 1233 (0,569) → 1224 (0,553) → 1134 (0,528) → 1125 (0,499) → 1116 (0,435); 234 (0,461) → 225 (0,432); and 135 (0,407) → 126 (0,369) → 117 (0,297).

The above sequences could be ordered in different ways. We have followed the rule of a “slow down” to include as many sequences in the main trends, as possible. With no exception, the above algorithm causes H₁ > H₂. To prove the inequality in a general case (for any 1 ≤ p ≤ q and n), we should show that

$$ {-}\left( {p/n} \right)\ln \left( {p/n} \right){-}\left( {q/n} \right)\ln \left( {q/n} \right) > {-}\left[ {\left( {p - 1} \right)/n} \right]\ln \left[ {\left( {p - 1} \right)/n} \right]{-}\left[ {\left( {q + 1} \right)/n} \right]\ln \left[ {\left( {q + 1} \right)/n} \right]. $$

If p → 1, then [(p − 1)/n] ln [(p − 1)/n] → 0. Hence, for p = 1 we get an obvious inequality (q + 1) (1 + 1/q)^q > 1. For 2 ≤ p ≤ q we should prove the inequality p^p/(p − 1)^p−1 < (q + 1)^q+1/q^q = f(q). Consider f(q) as a continuous function and use a logarithmic derivative df/dq = ln(1 + 1/q) × (q + 1)^q+1/q^q > 0. That is, f(q) grows with growing q = p, p + 1, p + 2, etc.

Let us show that the above inequality takes place even for the minimum argument q = p, i.e. p^p/(p − 1)^p−1 < (p + 1)^p+1/p^p or 1 < (p + 1)^p+1 (p − 1)^p−1/p^2p = f(p). Again, consider f(p) as a continuous function and use a logarithmic derivative df/dp = ln(1− 1/p²) × (p + 1)^p+1 (p − 1)^p−1/p^2p < 0. That is, f(p) drops with growing p = 2, 3, 4, etc. Indeed, f(2) = 1.6875, f(3) = 1.404…, f(4) = 1.287…, f(5) = 1.223…, f(6) = 1.182… Nevertheless, if p → ∞, then lim f(p) = lim (p + 1)^p+1 (p − 1)^p−1/p^2p = lim (1 + 1/p)^p (1− 1/p)^p [1 + 2/(p − 1)] = e × e⁻¹ × 1 = 1. That is, f(p) tends to 1 from above, i.e. f(p) > 1 for any p. Thus, H₁ > H₂ for any 1 ≤ p ≤ q and n.

Appendix 2

Assume that a convex polyhedron exists with all the facets being different (i.e. of different number of edges). Let us consider its Schlegel diagram on a facet with a maximum number of edges (k-lateral facet, Fig. 4a). More precisely, let us consider how its corona (i.e. a set of facets touching it edge-to-edge) is built. After (k − 1)-, (k − 2)- … 4-, and 3-lateral facets being attached to k-lateral one in any order, 3 more edges are free. And we can conclude that our initial assumption is wrong. Obviously, in the above case, 3 same (i.e. of the same number of edges), or 2 and 1, or 3 different facets can be attached to them. As any (i.e. 3- to k-lateral) facet is used, 4 same, or 3 and 2, or 3 pairs of same facets will result on a polyhedron.

Fig. 4

The Schlegel diagram on k-lateral facet (a) and the limit convex 4-, 5-, and 6-acra (b)

Now, assume that not all k − 3 types of the facets are submitted in the corona. After the facets of each type being attached by one to k-lateral facet, more than 3 edges are free. To complete the corona, one should choose more than 3 facets from their less than before (k − 3) variety. Obviously, both reasons may not reduce the frequency of occurrence of the facets in the corona: 4 same, or 3 and 2, or 3 pairs of same facets. Finally, because of the duality, any convex n-acron has at least 4, or 3 and 2, or 3 pairs of vertices of same valences. The limit cases are: a tetrahedron, a trigonal dipyramid, and a 6-acron of mm2 s.p.g. (Figure 4b).